x^2 - y^2=16 and xy=16 so find out x+y

• x^2-y^2=16
  (x+y)(x-y)=16
  so 16 comes in following table
  1*16,2*8,4*4
  using 2*8 equation
  x+y=8 and x-y=2 so x is 5 or 3 and y will be 3 or 5
  so answer is 8...
  not sure check it out
  
  
• 10 years agoHelpfull: Yes(15) No(43)
• sum will be
  4(sqrt(5)+1)/sqrt(2*sqrt(5)-1)
  
• 10 years agoHelpfull: Yes(4) No(8)
• 8
  x^2-y^2=16
  4*4 1*16,2*8
  x+y=8 and x-y=2
  so ans 8
  
• 10 years agoHelpfull: Yes(3) No(9)
• is d ans 16?
  
• 10 years agoHelpfull: Yes(0) No(2)
• answer is 8
  
• 8 years agoHelpfull: Yes(0) No(4)
• x^2 - y^2 = (x+y)(x-y) = 16;
  xy = 16;
  16 = 1 x 16 = 8 x 2 = 4 x 4;
  So we have to assume x and y such that the value of xy remains 16.
  If we take { (x+y) = 8 } and { (x-y) = 2 },
  8 = 1 x 8 = 4 x 4 = 2 x 4;
  If we take x=4 and y=4 then,
  x+y = 8.
  Basically, we have to prove x+y = 8 both sides.
  
  Final Answer is 8.
• 7 years agoHelpfull: Yes(0) No(5)
• x^2-y^2=16
  (x+y)(x-y)=16
  
  (x+y)^2(x-y)^2=256
  (x+y)^2{(x+y)^2-4xy}=256
  (x+y)^4-64(x+y)^2=256
  a^2-64a=256 //(x+y)^2=a//
  a=32+16(sqrt(5))
  x+y=8.24(app)
  x-y=1.94(app)
  x=5.09
  y=3.14
  x!=5 && y!=3 if so then xy!=16
  x!=4 && y!=4 if so then x^2 - y^2==0
• 5 years agoHelpfull: Yes(0) No(1)